Question

Let v = (v1, v2) be a vector in r2. show that (v2, −v1) is orthogonal to v, and use this fact to find two unit vectors orthogonal to the given vector. v = (9, 40)

Answer 1

Because v1 x v2 + v2 x (-v1) = v1 x v2 - v2 x v1 = v1 x v2 - v1 x v2 = 0, vector (v1,v2) is orthogonal to (v2,-v1);
Let (a,b) be an unit vector orthogonal to (9,40);
So, a^2 + b^2 = 1 and a x 9 + b x 40 = 0;
Then, a = - 40 x b / 9;
Finally, 160 x b^2 / 81 + b^2 = 1;
160 x b^2 + 81 x b^2 = 81;
241 x b^2 = 81;
b^2 = 81 / 241;
b = + or - 9/$\sqrt{241}$
a = + or - 40/$\sqrt{241}$
We have (+40/$\sqrt{241}$;9/$\sqrt{241}$) and (-40/$\sqrt{241}$;-9/$\sqrt{241}$);

Answer 2

Answer:

$\hat{u_1}=(\frac{40}{41},-\frac{9}{40})$ and $\hat{u_2}=(\frac{-40}{41},\frac{9}{41})$

Step-by-step explanation:

We are given that $v=(v_1,v_2)$ be a vector in $R^2$.

We have to show that $(v_2,-v_1)$ is orthogonal to v.We have to find the two unit vector orthogonal to the given vector v=(9,40).

Orthogonal vectors:If two vectors u and v are orthogonal then

$u\cdot v=u_1v_2+u_2v_1=0$

Using this condition

$(v_1,v_2)\cdot (v_2,-v_1)$

$=v_1v_2-v_2v_1=0$

Hence, $(v_2,-v_1)$ is orthogonal to v.

By using this fact then, we get

$u_1=$(40,-9) is orthogonal to v=(9,40)

Unit vector=$\frac{\vec{a}}{\mid a\mid}$

$\mid u_1\mid=\sqrt{(40)^2+(-9)^2}=41$

$\hat{u_1}=(\frac{40}{41},-\frac{9}{40})$

$u_2=(-40,9)$ is orthogonal to v=(9,40)

$\mid u_2\mid=\sqrt{(-40)^2+9^2}=41$

$\hat{u_2}=(\frac{-40}{41},\frac{9}{41})$

The two unit vectors which are orthogonal to given vector v=(9,40) are given by

$\hat{u_1}=(\frac{40}{41},-\frac{9}{40})$ and $\hat{u_2}=(\frac{-40}{41},\frac{9}{41})$